Download Assignment with Solved Problems - Differentiable Manifolds | MATH 6510 and more Assignments Mathematics in PDF only on Docsity!
Math 6510
Homework 5^1
C´esar Lozano
Problem 1 Given X 1 ,... , Xn ∈ Γ(T M ) and fi, gk ∈ C∞(M ) for 1 ≤ i, k ≤ n, let Y, Z ∈ Γ(T M ) be given by
Y (p) =
∑^ n
i
fi(p)Xi(p), and Z(p) =
∑^ n
k
gk(p)Xk(p).
Find an expression to Y, Z.
Solution. In order to find an expression to the field [Y, Z] in coordinates, we only need to make a computation. This one will be done keeping in mind that the Lie bracket is linear in each entry, i.e., [X + αY ; Z] = [X; Z] + α[Y ; Z]. As a result
[Y, Z] = [f 1 X 1 + · · · + fnXn; Z] = [f 1 X 1 + · · · + fnXn; g 1 X 1 + · · · + gnXn] =
i,j
[fiXi; gj Xj ]
Next, to deal with the last equation, let us see the following Lemma. [f X; gY ] = f g[X; Y ] + f LX gY − gLY f X. Solution. [f X, gY ] = f X(gY ) − gY (f X) = f gXY + f LX gY − gf Y X − gLY (f )X = f g[X; Y ] + f LX gY − gLY f X.
Consequently, doing the substitution, we get that Y ; Z has following expression ∑
ij
fi(p)gj (p)Xi; Xj + fi(p)(LXi gj )(p)Xj (p) − gj (p)(LXj fi)(p)Xi(p)
This completes the proof.
(^1) October 23 ,2008.
Problem 2 Compute [ (^) ∂x∂ ; (^) ∂y∂ + x (^) ∂z∂ ]
Solution. Consider the last formula where here f (x) = 1 and g(x) = x. As a result [ ∂ ∂x ,^
∂ ∂y +^ x^
∂ ∂z
]
= [ (^) ∂x∂ , (^) ∂y∂ ] + [ (^) ∂x∂ , x (^) ∂z∂ ]
= [ (^) ∂x∂ , (^) ∂z∂ ] + (^) ∂z∂ = (^) ∂z∂
Note [ (^) ∂x∂ , (^) ∂z∂ ] = [ (^) ∂x∂ , (^) ∂y∂ ] = 0 because the partial derivatives commute each other. This completes the proof.
Problem 3 If Γ ≤ Dif f (M n) acts freely and properly discontinuously on M , prove that the quotient map π : M n^ → M n/Γ is smooth.
Solution. For each p ∈ M n^ choose a chart (U, φ) such that both p ∈ U where φ(V ) ⊂ U and U ∪ gU = ∅ for g 6 = 1. Observe π|U is injective. Now note that
{ψ = φ ◦ π} : V → ΓM,
as well as the set V ∈ Rn^ is a differentiable structure on M n. Thus let
q ∈ ψ 1 (V 1 ) ∩ ψ 2 (V 2 ).
Let πi the restriction of π to φi(Vi) for i = 1, 2. Then let r ∈ W ⊂ V 2 , such that π ◦ φ(W ) ⊂ ψ 1 (V 1 ) ∪ ψ 2 (V 2 ). Finally observe that p 2 = π− 2 1 (q). Let p 1 = π− 1 1 ◦ π 2 (p 2 ). Observe that p 1 and p 2 are equivalent in M , hence there is a g ∈ Γ such that p 2 = gp 1. Consequently the restriction π− 1 1 ◦ π 2 |φ(W ) coincides with the diffeomorphism g|φ 2 (W ) which proves the the map π is smooth.
Problem 4 Let
NR =
1 x z 0 1 y 0 0 1
(^) |x, y, z ∈ R
, NZ =
1 x z 0 1 y 0 0 1
(^) |x, y, z ∈ Z
Prove that NN\NR^ is a compact 3 -manifold.
Solution. Firstly, we are going to establish that this set is, in fact, a 3-manifold. In order to do so, we will show the action of NZ on the manifold NR is both free and properly discontinuous. Free. It is clear that we can see NR as the space R^3 then the action in this notation becomes
(x, y, z)
g −→ (x + X, y + Y, z + xY + Z),
f : M × I //
π
M × I
π
M × S^1 = Mφn
f
??
F
/ / Mφ Observe that Mφn = M × S^1. Moreover note that the map f goes down to the quotient Mφ. Therefore, we get a map F : M × S^1 −→ Mφ,
such that |F −^1 (y)| = n for all y ∈ Mφ as desired. This completes the proof.
Problem 6 Prove that if φ ∈ Dif f (M ) then the mapping torus Mφ has a 1 -dimensional foliation and a codimension 1 foliation.
Solution. We want to show that
Mφ = M × I/ ∼,
where the equivalence relation ∼ identifies (x, 0) ∼ (φ(x), 1) has a 1-dimensional foliation corresponding to the interval I and a codimension 1 foliation corresponding to M. To show this we only need to show there is a foliation locally when we glue the manifold M along it self. However, keep in mind φ is a diffeomorphism, then it is clear whatever the case may be an open in Mφ looks like U × I 0 where U ⊂ M an I 0 ⊂ I.
Problem 7 Prove that S^2 does not have a one-dimensional foliation.
Solution. In order to show this fact we are going to argue by contradiction. Suppose S^2 does have a one-dimensional foliation. Then single out a point x ∈ U where (U 0 , φ 0 ) is a chart on S^2. Observe we can find a vector field in such open set compatible with the foliation. In other words, there is a vector field X ∈ Γ(T S^2 ), which has the leaves of the foliation as an integral curves. Since the atlas {Uα, φα} is compatible with the foliation consequently we can continue the vector field through the overlap to other chart U ′, φ′. Moreover S^2 is a compact set, therefore we only need a finitely many charts (U, φ). As a result we can continue the vector field X throughout S^2 by the foliation. Consequently, the Euler-characteristic vanishes
χ(S^2 ) = 0,
due to the field X vanishes nowhere. Here is the contradiction. Consequently, S^2 does not have a one-dimensional foliation.
Problem 8 Describe the Lie Algebra of
SO(2) =
cos(θ) − sin(θ) sin(θ) cos(θ)
|θ ∈ R
as subspace of gl 2 (R).
Solution. Observe SO(2) is basically a curve on GL 2 (R) which contains the identity element. Ex- plicitly,
α(t) =
cos(t) − sin(t) sin(t) cos(t)
As a result the Lie algebra is one-dimensional. Then we are going to take the derivative α′(0) in order to get the generator of such a Lie algebra. Accordingly we have
α′(0) =
Therefore, the Lie algebra has the expression {xα′(0)}|x ∈ R which observe this is the same of saying that so(2) =
A ∈ gl 2 (R)|A + At
where At^ denotes the transpose matrix of A. This completes the proof.
Problem 9 Find all the Lie subgroups of sl 2 (R) that contain
A =
1 x 0 1
|x ∈ R
Solution. Observe that such subgroups we are looking for must have Lie algebras. Such Lie algebras are contained in the Lie algebra of SL(R). Consequently, first we are going to count the subalgebras de Lie SL 2 (R). First note a subgroup A ⊂ H must have dimension either 1 or 2. If dimH = 1 since there are only two classes of homeomorphism of dimension 1. H must be either R or S^1. Since A ≈ R we have and A ⊂ H then H approxA. Therefore, A is the only 1-dimensional Lie subgroup containing to A. Next suppose dimH = 2. Observe that is generated by the following vectors
α 1 =
, α 2 =
, α 3 =
Thus taking derivative at the identity of the curve ( (^1) 0 1^ x ) the Lie subalgebra we are looking for must contain to the vector α 3 = ( 0 10 0 ). Let’s compute all the subspaces of sl 2 containing α 3. All of such subspaces,apart form one, can be parameterized by the real number t ∈ R as follows:^2 Wt = {x ∈ sl 2 |x = bα 3 + taα 2 + aα 1 }. (^2) Here δ is the slope, respecting to α 1 , of the line living in Wδ. Moreover ⊂ span{α 1 , α 2 }
The University of Utah
Department of Mathematics [email protected]
